Different forms of 2nd Bianchi's Identity
I have been struggling with relating two different forms of Bianchi's 2nd Identity: $dOmega = Omegawedgeomega-omegawedgeOmega$ and
$mathfrak{S}left{(nabla_{Z}R)(X,Y,Z)+R(T(X,Y),Z)W)right}=0$
here $R$ and $T$ represent curvature and torsion tensor of a connection $nabla$
$Omega$ and $omega$ represent curvature and connection form of $nabla$
How do we relate the exterior derivative of curvature form to the covariant derivative of curvature and torsion tensors?
I am able to prove them independently but I am not able to prove the equivalence.
I am pretty new to differential geometry and I apologize if this question is very easy.
Here are some of my thoughts:
I started with curvature form. Suppose $X_1,X_2,X_3,...X_n$ represent the moving frames and $theta^1, theta^2,...,theta_n$ represent the dual co frame. In the index notation the first form of bianchi's 2nd identity can be given as
$dOmega^{i}_{j} = Omega^i_kwedgeomega^k_j-omega^i_kwedgeOmega^k_j$
(using einstien's summation convention)
Relation between cuvature tensor and curvature form can be given as
$R(X_k,X_l)X_j = Omega^i_j(X_k,X_l)X_i$
in the index notation curvature tensor can be given as
$R(X_k,X_l)X_j = R^i_{jkl}X_i$
this gives us
$R^i_{jkl}X_i = Omega^i_j(X_k,X_l)X_i$ or
$Omega^i_{j} = frac{1}{2}sum_limits{k,l}R^i_{jkl}theta^kwedgetheta^l$ this implies
$dOmega^{i}_{j} =frac{1}{2}sum_limits{m,k,l} frac{partial{R^i_{j,k,l}}}{partial{X_m}}theta^mwedgetheta^kwedgetheta^l$
the covariant derivative of R in index notation can be given as:
$nabla R = sum_limits{h,j,k,l,i}R^i_{jkl;h}theta^hoplustheta^joplustheta^koplustheta^loplus X_i$
where $R^i_{jkl;h}$ can be given as
$R^i_{jkl;h} = frac{partial{R^i_{j,k,l}}}{partial{X_h}} + sumlimits_{v=1}^n R^v_{j,k,l}Gamma_{hv}^{i}-sum_limits{v=1}^{n}(R^i_{vkl}Gamma^{v}_{hj}+R^i_{kvl}Gamma^{v}_{hk}+R^i_{jkv}Gamma^{v}_{hl})$
where $Gamma$ can be given as
$nabla_{X_i}{X_j} = Gamma_{ij}^K X_k$
Also connection 1-form $omega^i_{j}$ can be given as:
$omega^i_j = Gamma^i_{kj}theta^k$
on one hand I have tensor product and on the other hand I have wedge product. At this point I am stuck and wonder if there is an alternate approach which doesn't involve coordinate representations. I am also not sure about my use of $X_i$'s as coordinate vector fields.
differential-geometry
asked Nov 28 '18 at 4:16
Akhil NagariyaAkhil Nagariya
112
add a comment |
I have been struggling with relating two different forms of Bianchi's 2nd Identity: $dOmega = Omegawedgeomega-omegawedgeOmega$ and
$mathfrak{S}left{(nabla_{Z}R)(X,Y,Z)+R(T(X,Y),Z)W)right}=0$
here $R$ and $T$ represent curvature and torsion tensor of a connection $nabla$
$Omega$ and $omega$ represent curvature and connection form of $nabla$
How do we relate the exterior derivative of curvature form to the covariant derivative of curvature and torsion tensors?
I am able to prove them independently but I am not able to prove the equivalence.
I am pretty new to differential geometry and I apologize if this question is very easy.
Here are some of my thoughts:
I started with curvature form. Suppose $X_1,X_2,X_3,...X_n$ represent the moving frames and $theta^1, theta^2,...,theta_n$ represent the dual co frame. In the index notation the first form of bianchi's 2nd identity can be given as
$dOmega^{i}_{j} = Omega^i_kwedgeomega^k_j-omega^i_kwedgeOmega^k_j$
(using einstien's summation convention)
Relation between cuvature tensor and curvature form can be given as
$R(X_k,X_l)X_j = Omega^i_j(X_k,X_l)X_i$
in the index notation curvature tensor can be given as
$R(X_k,X_l)X_j = R^i_{jkl}X_i$
this gives us
$R^i_{jkl}X_i = Omega^i_j(X_k,X_l)X_i$ or
$Omega^i_{j} = frac{1}{2}sum_limits{k,l}R^i_{jkl}theta^kwedgetheta^l$ this implies
$dOmega^{i}_{j} =frac{1}{2}sum_limits{m,k,l} frac{partial{R^i_{j,k,l}}}{partial{X_m}}theta^mwedgetheta^kwedgetheta^l$
the covariant derivative of R in index notation can be given as:
$nabla R = sum_limits{h,j,k,l,i}R^i_{jkl;h}theta^hoplustheta^joplustheta^koplustheta^loplus X_i$
where $R^i_{jkl;h}$ can be given as
$R^i_{jkl;h} = frac{partial{R^i_{j,k,l}}}{partial{X_h}} + sumlimits_{v=1}^n R^v_{j,k,l}Gamma_{hv}^{i}-sum_limits{v=1}^{n}(R^i_{vkl}Gamma^{v}_{hj}+R^i_{kvl}Gamma^{v}_{hk}+R^i_{jkv}Gamma^{v}_{hl})$
where $Gamma$ can be given as
$nabla_{X_i}{X_j} = Gamma_{ij}^K X_k$
Also connection 1-form $omega^i_{j}$ can be given as:
$omega^i_j = Gamma^i_{kj}theta^k$
on one hand I have tensor product and on the other hand I have wedge product. At this point I am stuck and wonder if there is an alternate approach which doesn't involve coordinate representations. I am also not sure about my use of $X_i$'s as coordinate vector fields.
differential-geometry
asked Nov 28 '18 at 4:16
Akhil NagariyaAkhil Nagariya
112
What does $omega$ and $Omega$ represent?
– Matt
Nov 28 '18 at 8:10
Hi @Matt I have added some details about the question
– Akhil Nagariya
Nov 29 '18 at 7:11
add a comment |
I have been struggling with relating two different forms of Bianchi's 2nd Identity: $dOmega = Omegawedgeomega-omegawedgeOmega$ and
$mathfrak{S}left{(nabla_{Z}R)(X,Y,Z)+R(T(X,Y),Z)W)right}=0$
here $R$ and $T$ represent curvature and torsion tensor of a connection $nabla$
$Omega$ and $omega$ represent curvature and connection form of $nabla$
How do we relate the exterior derivative of curvature form to the covariant derivative of curvature and torsion tensors?
I am able to prove them independently but I am not able to prove the equivalence.
I am pretty new to differential geometry and I apologize if this question is very easy.
Here are some of my thoughts:
I started with curvature form. Suppose $X_1,X_2,X_3,...X_n$ represent the moving frames and $theta^1, theta^2,...,theta_n$ represent the dual co frame. In the index notation the first form of bianchi's 2nd identity can be given as
$dOmega^{i}_{j} = Omega^i_kwedgeomega^k_j-omega^i_kwedgeOmega^k_j$
(using einstien's summation convention)
Relation between cuvature tensor and curvature form can be given as
$R(X_k,X_l)X_j = Omega^i_j(X_k,X_l)X_i$
in the index notation curvature tensor can be given as
$R(X_k,X_l)X_j = R^i_{jkl}X_i$
this gives us
$R^i_{jkl}X_i = Omega^i_j(X_k,X_l)X_i$ or
$Omega^i_{j} = frac{1}{2}sum_limits{k,l}R^i_{jkl}theta^kwedgetheta^l$ this implies
$dOmega^{i}_{j} =frac{1}{2}sum_limits{m,k,l} frac{partial{R^i_{j,k,l}}}{partial{X_m}}theta^mwedgetheta^kwedgetheta^l$
the covariant derivative of R in index notation can be given as:
$nabla R = sum_limits{h,j,k,l,i}R^i_{jkl;h}theta^hoplustheta^joplustheta^koplustheta^loplus X_i$
where $R^i_{jkl;h}$ can be given as
$R^i_{jkl;h} = frac{partial{R^i_{j,k,l}}}{partial{X_h}} + sumlimits_{v=1}^n R^v_{j,k,l}Gamma_{hv}^{i}-sum_limits{v=1}^{n}(R^i_{vkl}Gamma^{v}_{hj}+R^i_{kvl}Gamma^{v}_{hk}+R^i_{jkv}Gamma^{v}_{hl})$
where $Gamma$ can be given as
$nabla_{X_i}{X_j} = Gamma_{ij}^K X_k$
Also connection 1-form $omega^i_{j}$ can be given as:
$omega^i_j = Gamma^i_{kj}theta^k$
on one hand I have tensor product and on the other hand I have wedge product. At this point I am stuck and wonder if there is an alternate approach which doesn't involve coordinate representations. I am also not sure about my use of $X_i$'s as coordinate vector fields.
differential-geometry
asked Nov 28 '18 at 4:16
Akhil NagariyaAkhil Nagariya
112
I have been struggling with relating two different forms of Bianchi's 2nd Identity: $dOmega = Omegawedgeomega-omegawedgeOmega$ and
$mathfrak{S}left{(nabla_{Z}R)(X,Y,Z)+R(T(X,Y),Z)W)right}=0$
here $R$ and $T$ represent curvature and torsion tensor of a connection $nabla$
$Omega$ and $omega$ represent curvature and connection form of $nabla$
How do we relate the exterior derivative of curvature form to the covariant derivative of curvature and torsion tensors?
I am able to prove them independently but I am not able to prove the equivalence.
I am pretty new to differential geometry and I apologize if this question is very easy.
Here are some of my thoughts:
I started with curvature form. Suppose $X_1,X_2,X_3,...X_n$ represent the moving frames and $theta^1, theta^2,...,theta_n$ represent the dual co frame. In the index notation the first form of bianchi's 2nd identity can be given as
$dOmega^{i}_{j} = Omega^i_kwedgeomega^k_j-omega^i_kwedgeOmega^k_j$
(using einstien's summation convention)
Relation between cuvature tensor and curvature form can be given as
$R(X_k,X_l)X_j = Omega^i_j(X_k,X_l)X_i$
in the index notation curvature tensor can be given as
$R(X_k,X_l)X_j = R^i_{jkl}X_i$
this gives us
$R^i_{jkl}X_i = Omega^i_j(X_k,X_l)X_i$ or
$Omega^i_{j} = frac{1}{2}sum_limits{k,l}R^i_{jkl}theta^kwedgetheta^l$ this implies
$dOmega^{i}_{j} =frac{1}{2}sum_limits{m,k,l} frac{partial{R^i_{j,k,l}}}{partial{X_m}}theta^mwedgetheta^kwedgetheta^l$
the covariant derivative of R in index notation can be given as:
$nabla R = sum_limits{h,j,k,l,i}R^i_{jkl;h}theta^hoplustheta^joplustheta^koplustheta^loplus X_i$
where $R^i_{jkl;h}$ can be given as
$R^i_{jkl;h} = frac{partial{R^i_{j,k,l}}}{partial{X_h}} + sumlimits_{v=1}^n R^v_{j,k,l}Gamma_{hv}^{i}-sum_limits{v=1}^{n}(R^i_{vkl}Gamma^{v}_{hj}+R^i_{kvl}Gamma^{v}_{hk}+R^i_{jkv}Gamma^{v}_{hl})$
where $Gamma$ can be given as
$nabla_{X_i}{X_j} = Gamma_{ij}^K X_k$
Also connection 1-form $omega^i_{j}$ can be given as:
$omega^i_j = Gamma^i_{kj}theta^k$
on one hand I have tensor product and on the other hand I have wedge product. At this point I am stuck and wonder if there is an alternate approach which doesn't involve coordinate representations. I am also not sure about my use of $X_i$'s as coordinate vector fields.
differential-geometry
differential-geometry
asked Nov 28 '18 at 4:16
Akhil NagariyaAkhil Nagariya
112
asked Nov 28 '18 at 4:16
Akhil NagariyaAkhil Nagariya
112
edited Nov 29 '18 at 8:22
Akhil Nagariya
asked Nov 28 '18 at 4:16
Akhil NagariyaAkhil Nagariya
112
asked Nov 28 '18 at 4:16
Akhil NagariyaAkhil Nagariya
112
asked Nov 28 '18 at 4:16
Akhil NagariyaAkhil Nagariya
112
112
What does $omega$ and $Omega$ represent?
– Matt
Nov 28 '18 at 8:10
Hi @Matt I have added some details about the question
– Akhil Nagariya
Nov 29 '18 at 7:11
add a comment |
What does $omega$ and $Omega$ represent?
– Matt
Nov 28 '18 at 8:10
Hi @Matt I have added some details about the question
– Akhil Nagariya
Nov 29 '18 at 7:11
What does $omega$ and $Omega$ represent?
– Matt
Nov 28 '18 at 8:10
What does $omega$ and $Omega$ represent?
– Matt
Nov 28 '18 at 8:10
Hi @Matt I have added some details about the question
– Akhil Nagariya
Nov 29 '18 at 7:11
Hi @Matt I have added some details about the question
– Akhil Nagariya
Nov 29 '18 at 7:11
add a comment |
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What does $omega$ and $Omega$ represent?
– Matt
Nov 28 '18 at 8:10
Hi @Matt I have added some details about the question
– Akhil Nagariya
Nov 29 '18 at 7:11